12.Two ice skaters are initially at rest. The 78.2 kg male ice skater pushes his 48.5 kg female partner forward and away from hi

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12.Two ice skaters are initially at rest. The 78.2 kg male ice skater pushes his 48.5 kg female partner forward and away from hi

s body with a velocity of 8.46 m/s. What is the male skater's velocity as a result of the push

Physics

2 answers:

Keith_Richards [3.2K] · Answer 1 · 2026-04-08T06:55:58+03:00

The male skater's velocity is calculated to be -5.247 m/s. According to the principle of conservation of momentum, the total momentum before the collision equals that after. Since both skaters began at rest, the initial momentum is zero. Hence, we have 0 = mv + m'v', leading to -mv = m'v'. To isolate v: v = -(m'v'/m). Plugging in m = 78.2 kg, m' = 48.5 kg, and v' = 8.46 m/s into the equation provides v = -(48.5 × 8.46 / 78.2), which resolves to v = -5.247 m/s. The negative value indicates that the male skater moves in the reverse direction compared to the female skater.

Yuliya22 [3.3K] · Answer 2 · 2026-04-08T06:59:31+03:00

The male skater reaches a velocity of 13.71 m/s. According to the principle of conservation of momentum, m1u1 = (m1 + m2)u2, where m1 signifies the mass of the male skater at 78.2 kg, m2 is the mass of the female partner at 48.5 kg, u1 is the male skater's resulting velocity from the push, and u2 is the velocity imparted to the female skater, which was 8.46 m/s. Through the formula, we find u1 = [(78.2 + 48.5) × 8.46] ÷ 78.2, which calculates to 1071.882 ÷ 78.2 resulting in u1 = 13.71 m/s.